The number of roots of equation cosx+cos2x+cos3x=0 is (0≤x≤2π)
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Pranabamuni8085
09.08.2018
Math
Secondary School
+15 pts
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No of solutions of equation Cosx+cos2x+cos3x =0 in x belongs to 0 to pi
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abhi178
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Cosx + cos2x + cos3x = 0 like 
cosx + cos3x + cos2x = 0.
2cos(x + 3x)/2. cos(3x - x)/2 + cos2x = 0
[ from cosC + cosD = 2cos(C + D)/2.cos(C - D)/2]
2cos2x.cosx + cos2x = 0
cos2x(2cosx + 1) = 0
cos2x = 0, cosx = -1/2
when cos2x = 0 = cosπ/2
2x = 2nπ ± π/2 , where n is integers
x = nπ ± π/4
if n = 0, x = ± π/4 , but  so, x = π/4
if n = 1 , x = π ± π/4, x = 5π/4 , 3π/4 but x ≠ 5π/4
hence, x = π/4 and 3π/4 from cos2x = 0
when cosx = -1/2 = cos2π/3
x = 2nπ ± 2π/3
for  only x = 2π/3 is solution
hence, there are three solutions .e.g., x = π/4, 3π/4 and 2π/3
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